# Chapter17 - 17.1 Model For a gas the thermal energy is the...

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17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, E th = K micro . Solve: The number of atoms is N M m = = × = × 0 6 64 10 3 01 10 27 23 .0020 kg kg . . Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m = = × ( ) = × 4 6 64 10 27 u 4 1.661 10 kg kg 27 . The average kinetic energy of each atom is K mv avg avg 2 kg 700 m s = = × ( ) ( ) 1 2 1 2 27 2 6 64 10 . = 1.63 × 10 21 J Thus the thermal energy of the gas is E K NK th micro avg J = = = × ( ) × ( ) 3 01 10 1 63 10 23 21 . . = 490 J

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17.2. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. Solve: Oxygen atoms have an atomic mass number A = 16, so the mass of each molecule is m = 32 u = 32(1.661 × 10 27 kg) = 5.32 × 10 26 kg The number of molecules in the gas is N M m = = × 0 0080 26 . kg 5.32 10 kg = 1.505 × 10 23 The thermal energy is E NK N mv th avg avg 2 = = ( ) 1 2 = = ( ) × ( ) × ( ) v E Nm avg th 23 1700 J 1.505 10 kg 2 2 5 32 10 26 . = 650 m/s
17.3. Model: The work done on a gas is the negative of the area under the pV curve. Visualize: Please refer to Figure Ex17.3. The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is W pdV pV = − = − ( ) = − − ( ) ( ) ( ) = × ( ) × ( ) = area under the curve 200 cm 200 kPa m Pa 40 J 3 3 200 10 2 0 10 6 5 . Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environment does positive work on the gas to compress it.

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17.4. Model: The work done on a gas is the negative of the area under the pV curve. Visualize: Please refer to Figure Ex17.4. The gas is expanding, so we expect the work to be negative. Solve: The area under the pV curve is the area of the rectangle and triangle. We have 200 10 200 10 200 10 200 10 6 3 1 2 6 3 × ( ) × ( ) + × ( ) × ( ) m Pa m Pa 3 3 = 60 J Thus, the work done on the gas is W = 60 J. Assess: The environment does negative work on the gas as it expands.
17.5. Visualize: Please refer to Figure Ex17.5. Solve: The work done on gas in an isobaric process is W p V p V V = − = − ( ) f i Substituting into this equation, 80 J Pa = − × ( ) ( ) 200 10 3 3 1 1 V V = × = V i 3 3 m cm 2 0 10 200 4 . Assess: The work done to compress a gas is positive.

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17.6. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve: (a) Since the pressure ( p i = p f = p ) is constant the work done is W p V p V V nRT V V V on gas f i i i f i 3 3 3 0.10 mol 8.31 J mol K 573 K cm cm cm J = − = − = − ( ) = − ( ) ( ) ( ) ( ) = ( ) 1000 2000 2000 238 (b) For compression at a constant temperature, W nRT V V on gas f i 3 3 0.10 mol 8.31 J mol K 573 K m m 330 J = − ( ) = − ( ) ( ) ( ) × × = ln ln 1000 10 2000 10 6 6 (c) For the isobaric case, p nRT V = = × i i Pa 2 38 10 5 . For the isothermal case, p i Pa = × 2 38 10 5 . and the final pressure is p nRT V f f f Pa = = × 4 76 10 5 .
17.7. Visualize: Solve: Because W pdV = − and this is an isochoric process, W = 0 J. The final point is on a higher isotherm than the initial point, so T f > T i . Heat energy is thus transferred into the gas ( Q > 0) and the thermal energy of the gas increases ( E th f > E th i ) as the temperature increases.

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17.8.
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